where ##x## is the equilibrium concentration of either ##"Ag"^(+)## or ##"I"^(-)## in solution.

This means our equilibrium expression looks like this:

##K_"sp" = ["Ag"^(+)]["I"^(-)]##

##stackrel(K_"sp")overbrace(8.3xx10^(-17)) = x^2##

##=> color(green)(x = 9.1xx10^(-9))## ##color(green)("M")##

Since ##x## is the concentration of ##"Ag"^(+)## or ##"I"^(-)## in solution, and there is a ##1:1## molar ratio of ##"AgI"## to either of these species...

The molar solubility of silver iodide is ##color(blue)(9.1xx10^(-9))## ##color(blue)("M")##, or ##color(blue)("mol/L")##.

The physical interpretation of this is that silver iodide doesn't dissociate very much. It also happens to form a yellow precipitate in solution if there isn't enough water, demonstrating its poor solubility.